AS June 2019 Paper 1 Q7
7 A transformation A is represented by the matrix \(\mathbf{A}\) where \(\mathbf{A} = \begin{pmatrix} -1 & x & 2 \\ 7 - x & -6 & 1 \\ 5 & -5x & 2x \end{pmatrix}\).
The tetrahedron \(H\) has vertices at \(O\), \(P\), \(Q\) and \(R\). The volume of \(H\) is 6 units.
\(P'\), \(Q'\), \(R'\) and \(H'\) are the images of \(P\), \(Q\), \(R\) and \(H\) under A.
- find the volume of \(H'\),
- determine whether A preserves the orientation of \(H\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(x = 5 \Rightarrow \det\mathbf{A} = 2 \times 5^3 - 4 \times 5^2 - 58 \times 5 + 60\) \(= 250 - 100 - 290 + 60 = -80\) | B1 | 1.1 |
| So vol of \(H' = 6 \times 80 = 480\) cao | B1 | 1.1 |
| ...and A does not preserve the orientation because \(\det\mathbf{A} \lt 0\). | E1 ft | 2.4 |
| [3] |
Notes
B1: (1st) Could see e.g.-
\(\begin{vmatrix} -1 & 5 & 2 \\ 2 & -6 & 1 \\ 5 & -25 & 10 \end{vmatrix}\)
\(= -1(-60 + 25) - 5(20 - 5) + 2(-50 + 30)\)
\(= 35 - 75 - 40 = -80\)
B1: (2nd) not \(-480\)
E1 ft: Follow through on the sign of their determinant
If positive then “A does preserve the orientation because \(\det\mathbf{A} \gt 0\)”
| Scheme | Marks | AO |
|---|---|---|
| Image coplanar \(\Rightarrow \det\mathbf{A} = 0\) soi | B1 | 2.2a |
| \(\det\mathbf{A} = -1(-6 \times 2x + 5x) - x(2x(7 - x) - 5)\) \(+ 2(-5x(7 - x) + 6 \times 5)\) | M1 | 3.1a |
| \(2x^3 - 4x^2 - 58x + 60\) | A1 | 1.1 |
| \(1, 6, -5\) | A1 | 1.1 |
| [4] |
Notes
M1: Attempt to expand determinant
A1: (1st) Don’t need “=0” here
A1: (2nd) BC. All three
No working required for roots